The Scenario
A tall flagpole is held upright by a taut guide wire. The wire makes an angle $\theta$ with the horizontal ground. From the right-angled triangle formed by the pole, the ground and the wire, you already know three ratios:
$$\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}=\dfrac{h}{L}, \qquad \cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{d}{L}, \qquad \tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}=\dfrac{h}{d}.$$
An engineer needs to know the exact opposite: how many metres of wire are needed per metre of pole height — that is, the ratio $\dfrac{L}{h}=\dfrac{\text{hypotenuse}}{\text{opposite}}$. This is simply the flip of $\sin\theta$.
Review · Activating Prior Knowledge
Quick check — pick the correct statement in each question before we meet today's "flipped" cousins.
In a right-angled triangle, $\sin\theta=\dfrac{\text{opp}}{\text{hyp}}$, $\cos\theta=\dfrac{\text{adj}}{\text{hyp}}$ and $\tan\theta=\dfrac{\text{opp}}{\text{adj}}$. Every option looks complete — check the ratio you trust first. A swaps $\cos$ (that is really $\sec\theta$), B swaps $\sin$, C swaps $\tan$ (that is really $\cot\theta$). Only D has all three correct.
$\sin 30^{\circ}=\dfrac{1}{2}$, $\cos 30^{\circ}=\dfrac{\sqrt{3}}{2}$ and $\sin 45^{\circ}=\dfrac{\sqrt{2}}{2}$. These exact values return in today's reciprocal forms: for example $\cosec 30^{\circ}=2$ and $\sec 30^{\circ}=\dfrac{2\sqrt{3}}{3}$.
The Question
We already have names for $\dfrac{\text{opp}}{\text{hyp}}$, $\dfrac{\text{adj}}{\text{hyp}}$ and $\dfrac{\text{opp}}{\text{adj}}$. But what about the three flipped ratios $\dfrac{\text{hyp}}{\text{opp}}$, $\dfrac{\text{hyp}}{\text{adj}}$ and $\dfrac{\text{adj}}{\text{opp}}$? Are they just "one divided by something", or do they deserve their own names and properties?
💡 The Bridge (click to reveal)
Each flipped ratio is simply the reciprocal of a familiar one, and each gets its own name:
$$\sec\theta=\dfrac{1}{\cos\theta}, \qquad \cosec\theta=\dfrac{1}{\sin\theta}, \qquad \cot\theta=\dfrac{1}{\tan\theta}.$$
Today we learn these three reciprocal trigonometric functions and the graphs and identities built on them.
Red dashed lines mark vertical asymptotes (at $\theta = 0^\circ$ and $360^\circ$ the curve simply runs off the edge of the window). The faint dashed curve is the parent graph being reciprocated (cos θ or sin θ); it is omitted for cot θ.
| Function | Domain | Range |
|---|---|---|
| sec θ | $\theta \neq 90°, 270°, \ldots$ (i.e. $\cos\theta \neq 0$) | $(-\infty,\, -1] \cup [1,\, +\infty)$ |
| cosec θ | $\theta \neq 0°, 180°, 360°, \ldots$ (i.e. $\sin\theta \neq 0$) | $(-\infty,\, -1] \cup [1,\, +\infty)$ |
| cot θ | $\theta \neq 0°, 180°, 360°, \ldots$ (i.e. $\sin\theta \neq 0$) | $\mathbb{R}$ (all real numbers) |
Whole-class game (no time limit per round, up to 9 rounds): Designed for the projector screen. A detective steps up, types a nickname, then turns away from the screen. A classmate (or the teacher) secretly clicks one of the six graphs. The detective turns back and races the clock: click questions to ask about the mystery graph, then click the graph they believe is secret. Fewer questions and a faster solve = higher score! After each round: play another round with a new detective, or finish and hold the awards ceremony.
A new detective steps up!
Scoring: the round starts worth 30 pts and drops by 4 pts for every question asked (minimum 5 pts). Each wrong guess costs 5 pts. The clock counts up from 0 the moment you begin asking — no time limit. The faster you crack the case, the bigger your time bonus: +1 pt per second saved, up to 100 pts (solve in 0–1 s for the full 100; past 100 s the bonus runs out, but the base points always stand). There is no timeout penalty — keep asking until you are sure! Ask sharp questions, be quick, and trust your reasoning!
Discussion: Which single question eliminated the most graphs today? Which vocabulary — asymptote, period, turning point, range band — helped you ask sharper questions?
Individual activity (6 min): Now close your eyes and picture one of the curves you have just met. Choose $y=\cosec\theta$, $y=\sec\theta$ or $y=\cot\theta$, sketch one full period from $\theta = 0°$ to $\theta = 360°$ on the canvas, then check your sketch against the true graph and earn your match score. Stuck? Turn on the scaffold to see the dashed base trig graph as a guide.
Find the following giving your answers as fractions or in surd form. You should not need your calculator.
(i) $\cot 135°$ (ii) $\sec 150°$ (iii) $\cosec 240°$
▼ Reveal Solution
(i) $\cot 135°$: Q2, ref $= 45°$. $\cos 135° = -\dfrac{\sqrt{2}}{2}$, $\sin 135° = \dfrac{\sqrt{2}}{2}$. So $\cot 135° = \dfrac{\cos 135°}{\sin 135°} = \dfrac{-\sqrt{2}/2}{\sqrt{2}/2} = -1$.
(ii) $\sec 150°$: Q2, ref $= 30°$. $\cos 150° = -\dfrac{\sqrt{3}}{2}$. So $\sec 150° = \dfrac{1}{-\sqrt{3}/2} = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}$.
(iii) $\cosec 240°$: Q3, ref $= 60°$. $\sin 240° = -\dfrac{\sqrt{3}}{2}$. So $\cosec 240° = \dfrac{1}{-\sqrt{3}/2} = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}$.
(i) $-1$ (ii) $-\dfrac{2\sqrt{3}}{3}$ (iii) $-\dfrac{2\sqrt{3}}{3}$
Solve the following equations for $0^\circ \leqslant x \leqslant 360^\circ$.
(i) $\cosec x = 1$ (ii) $\sec x = 2$ (iii) $\cot x = 4$
▼ Reveal Solution
(i) $\cosec x = 1 \Rightarrow \sin x = 1$. The only angle in range with $\sin x = 1$ is $x = 90°$.
(ii) $\sec x = 2 \Rightarrow \cos x = \dfrac{1}{2}$. Reference angle $= 60°$. $\cos x > 0$ in Q1 and Q4: $x = 60°, \; 300°$.
(iii) $\cot x = 4 \Rightarrow \tan x = \dfrac{1}{4}$. Reference angle $= \arctan(0.25) \approx 14.0°$. $\tan x > 0$ in Q1 and Q3: $x \approx 14.0°, \; 194.0°$.
(i) $x = 90°$ (ii) $x = 60°, \; 300°$ (iii) $x \approx 14.0°, \; 194.0°$
Solve the following equations for $0^\circ \leqslant x \leqslant 360^\circ$.
(i) $\cos x = \sec x$ (ii) $\cosec x = \sec x$ (iii) $2 \sin x = 3 \cot x$
▼ Reveal Solution
(i) $\cos x = \sec x = \dfrac{1}{\cos x} \Rightarrow \cos^2 x = 1 \Rightarrow \cos x = \pm 1$: $x = 0°, 180°, 360°$.
(ii) $\cosec x = \sec x \Rightarrow \dfrac{1}{\sin x} = \dfrac{1}{\cos x} \Rightarrow \sin x = \cos x$ (both defined) $\Rightarrow \tan x = 1$: $x = 45°, 225°$.
$2\sin x = 3\cot x = \dfrac{3\cos x}{\sin x} \Rightarrow 2\sin^2 x = 3\cos x \Rightarrow 2(1 - \cos^2 x) = 3\cos x$.
So $2\cos^2 x + 3\cos x - 2 = 0 \Rightarrow (2\cos x - 1)(\cos x + 2) = 0$. $\cos x = \dfrac{1}{2}$: $x = 60°, 300°$; $\cos x = -2$ is impossible, so reject.
(i) $x = 0°, 180°, 360°$ (ii) $x = 45°, 225°$ (iii) $x = 60°, 300°$
Test your understanding of all four lesson objectives. Each question targets one objective.
By definition $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{1}{\sin\theta/\cos\theta} = \dfrac{\cos\theta}{\sin\theta}$. Option A confuses sec with cosec; B confuses cosec with sec; D multiplies instead of dividing.
$\sec 60° = \dfrac{1}{\cos 60°} = \dfrac{1}{1/2} = 2$. Note: $60°$ is in Q1 where $\cos$ is positive, so $\sec$ is also positive.
$\cosec x = -2 \Rightarrow \sin x = -\dfrac{1}{2}$. Reference angle $= 30°$. $\sin x < 0$ in Q3 and Q4, so $x = 180° + 30° = 210°$ and $x = 360° - 30° = 330°$.
Since $-1 \leqslant \cos\theta \leqslant 1$ (excluding $\cos\theta = 0$), taking reciprocals gives $\sec\theta \leqslant -1$ or $\sec\theta \geqslant 1$. The values between $-1$ and $1$ are never attained.
Solve the following equations for $0^\circ \leqslant x \leqslant 360^\circ$.
(iv) $\sec x = -3$ (v) $\cot x = -1$ (vi) $\cosec x = -2$
💡 Hint
Rewrite in terms of the base function: $\sec x = k \Rightarrow \cos x = \dfrac{1}{k}$, $\cot x = k \Rightarrow \tan x = \dfrac{1}{k}$, $\cosec x = k \Rightarrow \sin x = \dfrac{1}{k}$. A negative target means the angle lies in the two quadrants where that base function is negative.
▼ Show Solution
(iv) $\sec x = -3 \Rightarrow \cos x = -\dfrac{1}{3}$. Reference angle $= \arccos(1/3) \approx 70.5°$. $\cos x < 0$ in Q2 and Q3: $x \approx 180° - 70.5° = 109.5°, \; 180° + 70.5° = 250.5°$.
(v) $\cot x = -1 \Rightarrow \tan x = -1$. Reference angle $= 45°$. $\tan x < 0$ in Q2 and Q4: $x = 180° - 45° = 135°, \; 360° - 45° = 315°$.
(vi) $\cosec x = -2 \Rightarrow \sin x = -\dfrac{1}{2}$. Reference angle $= 30°$. $\sin x < 0$ in Q3 and Q4: $x = 180° + 30° = 210°, \; 360° - 30° = 330°$.
(iv) $x \approx 109.5°, \; 250.5°$ (v) $x = 135°, \; 315°$ (vi) $x = 210°, \; 330°$
Find the following giving your answers as fractions or in surd form. You should not need your calculator.
(iv) $\sec 210°$ (v) $\cot 270°$ (vi) $\cosec 225°$
💡 Hint
For each angle: (1) identify the quadrant, (2) find the reference angle, (3) evaluate the base function at the reference angle, (4) apply the quadrant sign, (5) take the reciprocal. Watch the axis angle $270°$: use $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$ directly.
▼ Show Solution
(iv) $\sec 210°$: Q3, ref $= 30°$. $\cos 210° = -\dfrac{\sqrt{3}}{2}$. So $\sec 210° = \dfrac{1}{-\sqrt{3}/2} = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}$.
(v) $\cot 270°$: $\cos 270° = 0$, $\sin 270° = -1$. So $\cot 270° = \dfrac{\cos 270°}{\sin 270°} = \dfrac{0}{-1} = 0$.
(vi) $\cosec 225°$: Q3, ref $= 45°$. $\sin 225° = -\dfrac{\sqrt{2}}{2}$. So $\cosec 225° = \dfrac{1}{-\sqrt{2}/2} = -\dfrac{2}{\sqrt{2}} = -\sqrt{2}$.
(iv) $-\dfrac{2\sqrt{3}}{3}$ (v) $0$ (vi) $-\sqrt{2}$
In triangle $ABC$, angle $A = 90^\circ$ and $\sec B = 2$.
(i) Find the angles $B$ and $C$.
(ii) Find $\tan B$.
(iii) Show that $1 + \tan^2 B = \sec^2 B$.
▼ Show Solution
$\sec B = 2 \Rightarrow \cos B = \dfrac{1}{2}$. Since $B$ is acute (angle in a right-angled triangle), $B = 60°$.
Angle $C = 90° - 60° = 30°$ (angles in a triangle sum to $180°$).
$B = 60°, \; C = 30°$
$\tan B = \tan 60° = \sqrt{3}$
LHS: $1 + \tan^2 B = 1 + (\sqrt{3})^2 = 1 + 3 = 4$.
RHS: $\sec^2 B = 2^2 = 4$.
LHS $=$ RHS $= 4$, so $1 + \tan^2 B = \sec^2 B$. $\checkmark$
In triangle $LMN$, angle $M = 90^\circ$ and $\cot N = 1$.
(i) Find the angles $L$ and $N$.
(ii) Find $\sec L$, $\cosec L$ and $\tan L$.
(iii) Show that $1 + \tan^2 L = \sec^2 L$.
▼ Show Solution
$\cot N = 1 \Rightarrow \tan N = 1$. Since $N$ is acute, $N = 45°$.
Angle $L = 90° - 45° = 45°$.
$L = 45°, \; N = 45°$
$\cos 45° = \dfrac{\sqrt{2}}{2}$, so $\sec L = \dfrac{1}{\sqrt{2}/2} = \dfrac{2}{\sqrt{2}} = \sqrt{2}$.
$\sin 45° = \dfrac{\sqrt{2}}{2}$, so $\cosec L = \dfrac{1}{\sqrt{2}/2} = \sqrt{2}$.
$\tan L = \tan 45° = 1$
LHS: $1 + \tan^2 L = 1 + 1^2 = 2$.
RHS: $\sec^2 L = (\sqrt{2})^2 = 2$.
LHS $=$ RHS $= 2$, so $1 + \tan^2 L = \sec^2 L$. $\checkmark$
Malini is 1.5 m tall. At 8 pm one evening her shadow is 6 m long. Given that the angle of elevation of the sun at that moment is $\alpha$
(i) show that $\cot \alpha = 4$
(ii) find $\alpha$
💡 Hint
Draw a right-angled triangle: Malini's height is the opposite side to $\alpha$, and the shadow length is the adjacent side. Then $\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}}$, and $\cot\alpha$ is its reciprocal.
▼ Show Solution
In the right-angled triangle formed by Malini and her shadow:
opposite $= 1.5$ m (height), adjacent $= 6$ m (shadow).
$\tan\alpha = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{1.5}{6} = \dfrac{1}{4}$.
Therefore $\cot\alpha = \dfrac{1}{\tan\alpha} = \dfrac{4}{1} = 4$. $\checkmark$
$\cot\alpha = 4 \Rightarrow \tan\alpha = \dfrac{1}{4}$.
$\alpha = \arctan\!\left(\dfrac{1}{4}\right) \approx 14.0°$ (to 1 d.p.).
$\alpha \approx 14.0°$
Solve the following equations for $0^\circ \leqslant x \leqslant 360^\circ$.
(iv) $\cosec^2 x + \cot^2 x = 2$ (v) $3 \sec^2 x - 10 \tan x = 0$ (vi) $1 + \cot^2 x = 2 \tan^2 x$
💡 Hint
Use the Pythagorean identities $\cosec^2 x = 1 + \cot^2 x$ and $\sec^2 x = 1 + \tan^2 x$ to reach a quadratic in one function. For part (vi), substitute $t = \tan x$ (so $\cot x = \dfrac{1}{t}$). Reject impossible cases such as $\tan^2 x < 0$.
▼ Show Solution
(iv) $\cosec^2 x + \cot^2 x = 2 \Rightarrow (1 + \cot^2 x) + \cot^2 x = 2 \Rightarrow 2\cot^2 x = 1 \Rightarrow \cot x = \pm\dfrac{1}{\sqrt{2}}$, i.e. $\tan x = \pm\sqrt{2}$. Reference angle $\approx 54.7°$: $x \approx 54.7°, 125.3°, 234.7°, 305.3°$.
(v) $3\sec^2 x - 10\tan x = 0 \Rightarrow 3(1 + \tan^2 x) - 10\tan x = 0 \Rightarrow 3\tan^2 x - 10\tan x + 3 = 0 \Rightarrow (3\tan x - 1)(\tan x - 3) = 0$. $\tan x = \dfrac{1}{3}$ (ref. $\approx 18.4°$): $x \approx 18.4°, 198.4°$; $\tan x = 3$ (ref. $\approx 71.6°$): $x \approx 71.6°, 251.6°$.
(vi) $1 + \cot^2 x = 2\tan^2 x$. Let $t = \tan x$, so $\cot x = \dfrac{1}{t}$: $1 + \dfrac{1}{t^2} = 2t^2 \Rightarrow t^2 + 1 = 2t^4 \Rightarrow 2t^4 - t^2 - 1 = 0 \Rightarrow (2t^2 + 1)(t^2 - 1) = 0$. $t^2 = 1 \Rightarrow \tan x = \pm 1$ ($2t^2 + 1 = 0$ is impossible): $x = 45°, 135°, 225°, 315°$.
(iv) $x \approx 54.7°, 125.3°, 234.7°, 305.3°$ (v) $x \approx 18.4°, 71.6°, 198.4°, 251.6°$ (vi) $x = 45°, 135°, 225°, 315°$
Exit Ticket
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